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To prove for an open set, the inverse of a continuous function is continuous.

The proof is,

Let $F:\mathbb{R}\rightarrow\mathbb{R}$ be continuous. It is sufficient that $F^{-1}$ is the inverse of $F$ and $W\subseteq\mathbb{R}$ is open then $(F^{-1})^{-1}(W)$ is open. But

$(F^{-1})^{-1}(W)=\{x\in\mathbb{R}|\exists w\in W:F^{-1}(x)=w\}=\{x\in\mathbb{R}|\exists w\in W:x=F(w)\}=F(W)$

and so $(F^{-1})^{-1}(W)$ is open if and only if $F(W)$ is open. Since $F(W)$ is open $W$ is open and $F$ is continuous.

This is the proof, I am unsure if it is correct also is there any example that can prove this?

Vicky
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    This result is false, there's a bunch of counterexamples here: https://math.stackexchange.com/questions/68800/functions-which-are-continuous-but-not-bicontinuous – bitesizebo Oct 14 '18 at 01:32
  • You are correct that the preimage of a set $W$ under $F^{-1}$ (assuming it exists) equals the image set $F(W)$ but notice the definition says that a function $f$ is continuous if and only if $f^{-1}(W)$ is open whenever $W$ is open. You seem to be committing some error of either confusing the direction of the implication in this definition or assuming without reason that $F$ is an open map. – M A Pelto Oct 14 '18 at 01:34
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    You have no reason to believe that $F(W)$ will be open when you assume that $F$ is continuous (indeed, this is equivalent to what you are attempting to prove). That would mean that $F^{-1}(W)$ would be guaranteed open. – Randall Oct 14 '18 at 01:46
  • You should also explicitly write down the assumption that $F$ is bijective, otherwise it doesn't make sense. – Randall Oct 14 '18 at 01:48
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    @bitesizebo if OP is asking about continuous bijections on $\mathbb{R}$ with the usual topology, this result is actually true. Note that none of the counterexamples in the link are counterexamples to this claim. – Randall Oct 14 '18 at 01:54
  • to OP: unless you know about compactness, this result is a little tough to get. (There may be other, "friendlier" approaches that I do not know about.) No matter what, I'm pretty sure any proof would need to make use of the (necessarily true) monotonicity of such a function. – Randall Oct 14 '18 at 01:57
  • Informal cliffs: Any continuous bijection with compact domain and Hausdorff codomain is a closed map. Every metric space is Hausdorff, and $\mathbb{R}$ with the standard topology is locally compact Hausdorff. So if $f:\mathbb{R} \to \mathbb{R}$ is a continuous bijection, then $f^{-1}$ is continuous at each point $y \in \mathbb{R}$ since $y$ has a compact neighborhood $K_y$ and $f{\restriction_{K_y}}$ is a closed map. – M A Pelto Oct 14 '18 at 02:45
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    Or, $f([a,b])$ must be compact and connected, hence it must be $[c,d]$. But $f$ is monotone (assume increasing), so $f(a)=c$ and $f(b)=d$. Delete ends, so $f(a,b) =(c,d)$. Hence $f$ maps all open sets to open sets. – Randall Oct 14 '18 at 03:01

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